Question

x+2squarerootx-3=0. solve for x?

Answer 1

\[x+2\sqrt{x}-3=0\]
\[2\sqrt{x}=3-x\]
square to get
\[4x=(3-x)^2\]
\[4x=9-6x+x^2\]
\[x^2-10x+9=0\] solve for x get
\[(x-1)(x-10)=0\]
\[x=1\] or \[x=10\] butyou are not done here because you have to check your answers.

Answer 2

ok that was way wrong
\[x^2-10x+9=0\]
\[(x-1)(x-9)=0\]
\[x=1\] or
\[x=9\] now it is right

Answer 3

now check because there is no guarantee they work
\[1+2\sqrt{1}-3=0\]
\[1+2-3=0\] that one works
\[9+2\sqrt{9}-3=0\]
\[9+3-3=0\] that one does not work. only solution is
\[x=1\]